What is the solution for y^'=5y-ty^2 ?

The solution for y^'=5y-ty^2 is y=(25e^{5t})/(5e^{5t)t-e^{5t}+c_{1}}

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y^'=5y-ty^2

y^{^{'}} = 5 y - t y^{2}

y = \frac{25 e ^{5 t}}{5 e ^{5 t} t - e ^{5 t} + c _{1}}

Solution steps

y^{'} (t) = 5 y - t y^{2}

Solve Bernoulli ODE:

y = \frac{25 e ^{5 t}}{5 e ^{5 t} t - e ^{5 t} + c _{1}}

y = \frac{25 e ^{5 t}}{5 e ^{5 t} t - e ^{5 t} + c _{1}}

\int (3 x^{2} - 5 x^{\frac{3}{2}} + 6 x) d x

(2 x - 1) d x + (3 y + 7) d y = 0

x \to 4 lim (\frac{2}{x + 5})

\frac{d}{d y} (\frac{( z x - y ) ^{2}}{x})

X \frac{9}{6}, 7

What is the solution for y^'=5y-ty^2 ?
The solution for y^'=5y-ty^2 is y=(25e^{5t})/(5e^{5t)t-e^{5t}+c_{1}}